The sum of the infinite series $$1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,.....$$ is equal to :

<p>$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} +$ ...... (1)</p> <p>${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} +$ ...... (2)</p> <p>$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} +$ ........</p> <p>$$ \Rightarrow {{5S} \over 6} = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ....... (3)</p> <p>Now, multiplying both sides by ${1 \over 6}$, we get</p> <p>$$ \Rightarrow {{5S} \over {36}} = {1 \over 6} + {4 \over {{6^2}}} + {7 \over {{6^3}}} + {{10} \over {{6^4}}} + $$ ...... (4)</p> <p>Subtract equation (4) from equation (3), we get</p> <p>${{25} \over {36}}S = 1 + {3 \over 6} + {3 \over {{6^2}}} + {3 \over {{6^3}}} +$ ......</p> <p>$\Rightarrow {{25S} \over {36}} = 1 + {{{3 \over 5}} \over {1 - {1 \over 6}}}$</p> <p>$= 1 + {3 \over 6} \times {6 \over 5}$</p> <p>$= 1 + {3 \over 5} = {8 \over 5}$</p> <p>$\Rightarrow S = {8 \over 5} \times {{36} \over {25}} = {{288} \over {125}}$</p>